Solve for $x$ : $2x^2 - 6x - 8 = 0$
Solution: Dividing both sides by $2$ gives: $ x^2 {-3}x {-4} = 0 $ The coefficient on the $x$ term is $-3$ and the constant term is $-4$ , so we need to find two numbers that add up to $-3$ and multiply to $-4$ The two numbers $-4$ and $1$ satisfy both conditions: $ {-4} + {1} = {-3} $ $ {-4} \times {1} = {-4} $ $(x {-4}) (x + {1}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x -4) (x + 1) = 0$ $x - 4 = 0$ or $x + 1 = 0$ Thus, $x = 4$ and $x = -1$ are the solutions.